Let $p_n=\mathbb{P}(X_n=1\mid X_0=1)$. Conditioning on the previous state gives $p_{n+1}=(1-a)p_n+b(1-p_n)=b+(1-a-b)p_n$, with $p_0=1$. Its fixed point is $b/(a+b)$; subtracting it leaves a geometric recursion. Therefore

For $n=0$, the identity matrix gives the assertion. Suppose it holds at $n$. Conditioning on the state at time $n$ and using the Markov property givesThe nonnegative countable sum is justified by the law of total probability. Induction proves the $n$-step transition matrix is $P^n$.

Part (b) gives $V_i\sim U(0,1)$, so $\mathbb{E}{V}_i=1/2$ and $\mathrm{Var}(V_i)=1/12$. Independence yields $\mathbb{E}(S_n/n)=1/2$ and $\mathrm{Var}(S_n/n)=1/(12n)$. Apply part (a) to $Z=S_n/n-1/2$:This is a direct Chebyshev inequality for the sample mean.

The joint density of $X,Y$ is ${\lambda }^2{e}^{-\lambda (x+y)}$ for positive $x,y$. Inverting the transformation gives $x=uv$, $y=u(1-v)$, with absolute Jacobian determinant $u$ and domain $u>0$, $0<v<1$. The change of variables for probability densities givesThe joint density factors into the gamma density of $U$ and the uniform density of $V$ on $(0,1)$, so these two transformed variables are independent.

Pointwise $Z^2\ge a^2{1}_{\{|Z|\ge a\}}$. Taking expectations and dividing by $a^2>0$ proves the Markov inequality

The derivative of the pair's total angular momentum is $q_1\times F-q_2\times F=(q_1-q_2)\times F=0$, since $F$ is parallel to the separation vector. Therefore the combined angular momentum of particles one and two is conserved, despite the possible exchange between them individually.

The third particle is free, soThus particle three's angular momentum is conserved.

The Lagrangian is $L={\sum }_i{m}_i|{\dot{q}}_i{|}^2/2-f(r)$. Away from a collision, its equations of motion areSumming forces proves conservation of total linear momentum. The total mechanical energy $E={\sum }_i{m}_i|{\dot{q}}_i{|}^2/2+f(r)$ is also conserved: its derivative is $F\cdot ({\dot{q}}_1-{\dot{q}}_2)+f^{\prime }(r)\dot{r}=0$, because $\dot{r}=(q_1-q_2)\cdot ({\dot{q}}_1-{\dot{q}}_2)/r$.

For the first proposed angular momentum,which need not vanish. Thus particle one's angular momentum about the fixed origin is not generally conserved. Its force is central relative to particle two, not necessarily relative to that origin.

The constraint intersection is nonempty, closed and bounded, hence compact, so extrema exist. The two constraint gradients are $(x/2,2y,2z)$ and $(1,1,-1)$. If the first were $t$ times the second, the plane condition would give $2t+t/2+t/2=0$, so $t=0$ and all coordinates would vanish, contradicting the ellipsoid equation. Their rank is therefore two at every feasible point, justifying the Lagrange multiplier equations.

These equations areEliminating $\mu$ gives $1=\lambda (5x/2+2y)$ and $2=\lambda (2x+4y)$. The first equation makes $\lambda \ne 0$; subtracting twice the first from the second forces $x=0$. Hence $z=y$, and the ellipsoid gives $y=z=\pm 1/\sqrt{2}$. ThusAll candidates were found, and compactness ensures they include the extrema.

The full-rank constraint qualification gives a multiplier vector $\lambda \in {\mathbb{R}}^2$ such thatEquivalently the gradient of the objective lies in the span of the two constraint gradients. The point also satisfies $f(x_0)=0$.